Polynomials and Logarithms

Overview

This chapter covers two high-yield topics for CMI: polynomials and logarithms. Focus on definitions, core theorems, standard transformations, and short problem-solving patterns.

---

Chapter Contents

| # | Topic | What You'll Learn |
|---|---|---|
| 1 | Introduction to Polynomials | Degree, roots, factorization, Vieta, transformations |
| 2 | Logarithms and Their Properties | Definition, domain, laws, change of base, inequalities |

---

Learning Objectives

---

Part 1: Introduction to Polynomials

---

Key Concepts

1. Classification by Degree

• Degree $0$: constant, e.g. $5$
• Degree $1$: linear, e.g. $2x-3$
• Degree $2$: quadratic, e.g. $x^2-4x+4$
• Degree $3$: cubic, e.g. $x^3+2x^2-x+1$
• Degree $4$: quartic, e.g. $x^4-3x^2+2$

---

2. Basic Operations

Addition / subtraction: combine like terms.

Example
Let $P(x)=3x^3-2x+5$ and $Q(x)=x^3+4x^2-7$.

$P(x)+Q(x)=(3x^3-2x+5)+(x^3+4x^2-7)=4x^3+4x^2-2x-2$

Multiplication: distribute each term.

Example
Let $P(x)=x-2$ and $Q(x)=x^2+3x+1$.

P(x)Q(x) & = (x-2)(x^2+3x+1)
&=x^3+3x^2+x-2x^2-6x-2
&=x^3+x^2-5x-2

---

3. Division, Remainder Theorem, Factor Theorem

If a polynomial $P(x)$ is divided by a non-zero polynomial $D(x)$, then

Example
Find the remainder when $P(x)=x^3-2x^2+5x-1$ is divided by $(x-1)$.

So the remainder is $3$, and $(x-1)$ is not a factor.

---

4. Roots and Fundamental Theorem of Algebra

Multiplicity
A root $r$ has multiplicity $k$ if $(x-r)^k$ divides $P(x)$ but $(x-r)^{k+1}$ does not.

---

5. Vieta's Formulas

For

with roots $r_1,r_2,\dots,r_n$,

Quadratic example
For $P(x)=2x^2-6x+4$:

• sum of roots $=-\dfrac{-6}{2}=3$
• product of roots $=\dfrac{4}{2}=2$

---

6. Rational Root Theorem and Complex Conjugate Root Theorem

Example
Find all rational roots of $P(x)=2x^3+x^2-7x-6$.

Possible rational roots are

Now test:

So the rational roots are

---

7. Polynomial Transformations

If $r$ is a root of $P(x)$, then:

Roots of $P(x+c)$ are $r-c$

Roots of $x^nP(1/x)$ are reciprocals of the non-zero roots of $P(x)$

Example
If $f(x)$ has roots $r_1,r_2,r_3$, then roots of $f(x+2)$ are

If

then roots of $x^3f(1/x)$ are

---

8. Polynomial Functions and IVT

Polynomials are continuous everywhere.

Example
Show that $f(x)=x^3+x-1$ has a root in $(0,1)$.

Since $f(0)f(1)<0$, by IVT there is at least one root in $(0,1)$.

---

9. Integer Coefficient Property

This is very useful for existence / impossibility questions.

---

10. GCD of Polynomials

The GCD of two polynomials is the highest-degree polynomial dividing both.

Example
Find the GCD of

Factor:

So the GCD is

---

Common Mistakes

---

Practice Questions

:::question type="MCQ" question="Let $P(x)=3x^4-2x^3+ax^2-5x+7$. If $P(1)=10$, what is the value of $a$?" options=["A) 7","B) 0","C) -3","D) 1"] answer="A" hint="Substitute $x=1$." solution="

Given $P(1)=10$, so

Hence the correct option is A."
:::

:::question type="NAT" question="A polynomial $P(x)$ has roots $1,-2,$ and $3$. If the leading coefficient is $2$, what is the constant term of $P(x)$?" answer="12" hint="Use the product-of-roots formula." solution="For a cubic polynomial,

Now

So

Therefore the constant term is $12$."
:::

:::question type="MSQ" question="Let $f(x)$ be a polynomial of degree $4$ with real coefficients. Which of the following statements is/are always true?" options=["A) If $f(x)$ has one complex root $a+bi$ with $b \ne 0$, it must have at least one other complex root.","B) $f(x)$ must have at least two real roots.","C) The graph of $f(x)$ can intersect the x-axis at exactly three points.","D) If $f(x)$ has roots $r_1,r_2,r_3,r_4$, then the roots of $f(2x)$ are $r_1/2,r_2/2,r_3/2,r_4/2$."] answer="A,C,D" hint="Use conjugate roots and root transformations." solution="A) True. Non-real roots of real-coefficient polynomials occur in conjugate pairs.
B) False. Example: $x^4+1$ has no real roots.
C) True. Example: $(x-1)^2(x-2)(x-3)$ intersects the x-axis at exactly three distinct points.
D) True. If $f(r)=0$, then for $f(2x)=0$ we need $2x=r$, so $x=r/2$.

Hence the correct options are A, C, D."
:::

:::question type="SUB" question="Consider the polynomial $P(x)=x^3-3x^2+2x-1$. (a) Show that $P(x)$ has at least one real root between $x=2$ and $x=3$. (b) If $r_1,r_2,r_3$ are the roots of $P(x)$, find the value of $r_1r_2+r_1r_3+r_2r_3$." answer="a) Root exists in (2,3), b) 2" hint="For part (a), use IVT. For part (b), use Vieta." solution="(a)

Since $P(2)<0$ and $P(3)>0$, by IVT there is at least one real root in $(2,3)$.

(b)
For

the sum of pairwise products of roots is the coefficient of $x$ divided by the leading coefficient:

So the answer is:

• part (a): root exists in $(2,3)$


• part (b): $2$"

:::

:::question type="NAT" question="The polynomial $P(x)=x^4+ax^3+bx^2+cx+d$ has roots $1,2,3,$ and $4$. What is the coefficient $a$?" answer="-10" hint="Use the sum of roots." solution="For a monic quartic,

Here

So
"
:::

:::question type="MCQ" question="Which of the following describes the roots of $P(x)=x^4-16$?" options=["A) Four real distinct roots","B) Two real roots and two complex conjugate roots","C) Four non-real roots","D) Two real roots and two repeated real roots"] answer="B" hint="Factor completely." solution="

So the roots are $2,-2,2i,-2i$: two real roots and two complex conjugate roots.
Hence the correct option is B."
:::

:::question type="SUB" question="Does there exist a polynomial $P(x)$ with integer coefficients such that $P(0)=5$ and $P(2)=8$? Justify your answer." answer="No" hint="Use the divisibility property for integer-coefficient polynomials." solution="If $P(x)$ has integer coefficients, then for integers $a,b$,

Take $a=2$ and $b=0$. Then

must divide

But $2$ does not divide $3$.

Therefore, no such polynomial exists."
:::

---

Summary for Part 1

---

Part 2: Logarithms and Their Properties

---

Key Concepts

1. Definition and Constraints

Valid only when

• $a>0$


• $b>0$


• $b\ne 1$

---

2. Domain and Range

Example
Find the domain of $f(x)=\log_3(x^2-4x+3)$.

So the domain is

---

3. Laws of Logarithms

Example
If $f(xy)=f(x)+f(y)$ on positive reals and $f(8)=3$, find $f(1/4)$.

Since $8=2^3$,

Then

Also

So

---

4. Change of Base

Example
Simplify

Numerator:

So

Let $\log_8 32=y$. Then

---

5. Logarithmic Inequalities

Example
Check whether

Since

the right side is

Numerically,

So the inequality is true.

---

6. A Useful Algebraic Log Example

If $x^2+y^2=z^2$, then

can be simplified as follows.

Let

Then

So

Since $x^2+y^2=z^2$,

Hence

---

Common Mistakes

---

Practice Questions

:::question type="MCQ" question="What is the domain of the function $f(x)=\log_5(x^2-7x+10)$?" options=["A) $(-\infty,2)$","B) $(5,\infty)$","C) $(-\infty,2)\cup(5,\infty)$","D) $(2,5)$"] answer="C" hint="Require the argument to be positive." solution="We need

Factor:

This holds for

So the domain is

Hence the correct option is C."
:::

:::question type="NAT" question="If $\log_a b=2$ and $\log_b c=3$, what is the value of $\log_c a^2$?" answer="1/3" hint="Rewrite everything in powers of $a$." solution="From

and

Now

So the answer is
"
:::

:::question type="MSQ" question="Let $g$ be a function on the positive real numbers such that $g(x^y)=y\,g(x)$. If $g(27)=3$, which of the following statements is/are true?" options=["A) $g(3)=1$","B) $g(9)=2$","C) $g(1)=0$","D) $g(81)=4$"] answer="A,B,C,D" hint="Use $27=3^3$ and then apply the rule repeatedly." solution="Since $27=3^3$,

So

Hence A is true.

Then

So B is true.

Also

So C is true.

And

So D is true.

Hence all four options are correct."
:::

:::question type="SUB" question="Given $a,b,c$ are positive real numbers such that $a^x=b^y=c^z=K$, prove that $\frac1x+\frac1y+\frac1z=\log_K(abc)$." answer="Proof shows LHS = $\log_K(abc)$" hint="Convert $1/x,1/y,1/z$ into logarithms with base $K$." solution="Since

taking $\log_K$ gives

Similarly,

Therefore,

Hence proved."
:::

:::question type="MCQ" question="Which of the following statements is true?" options=["A) $\log_2 7<\log_3 7$","B) $\log_2 7>\log_3 7$","C) $\log_2 7=\log_3 7$","D) The relationship depends on the value of $7$"] answer="B" hint="Use change of base." solution="

Since $\ln 7>0$ and $\ln 2<\ln 3$, the fraction with denominator $\ln 2$ is larger. Hence

So the correct option is B."
:::

:::question type="NAT" question="Evaluate $3^{\log_3 5}+2^{\log_2 7}-5^{\log_5 2}$." answer="10" hint="Use $b^{\log_b x}=x$." solution="Using $b^{\log_b x}=x$:

So the expression is

Hence the answer is $10$."
:::

:::question type="MSQ" question="Which of the following expressions are equivalent to $\log(x^2y^3)$? Assume $x,y>0$." options=["A) $2\log x+3\log y$","B) $\log(x^2)+\log(y^3)$","C) $5\log(xy)$","D) $\log(x^2)\cdot\log(y^3)$"] answer="A,B" hint="Use product and power rules." solution="

So A and B are equivalent.

For C:

For D:
it is a product of two logarithms, not a single logarithm of a product.

Hence the correct options are A, B."
:::

:::question type="NAT" question="If $\log_2 x+\log_4 x+\log_8 x=11$, what is the value of $x$?" answer="64" hint="Write all logs in base $2$." solution="Let

Then

So



Thus
"
:::

---

Summary for Part 2

---

Chapter Review Questions

:::question type="MCQ" question="If one root of the quadratic polynomial $P(x)=x^2-5x+k$ is $\log_2 8$, what is the value of $k$?" options=["A) 6","B) 2","C) -6","D) -2"] answer="A" hint="First simplify the logarithm." solution="

Since $3$ is a root,

So



Hence the correct option is A."
:::

:::question type="NAT" question="If $2^{\log_8(x^2-1)}=3$, find the value of $x^2$." answer="28" hint="Set $y=\log_8(x^2-1)$." solution="Let

Then

Now

But from the definition of logarithm,

So
"
:::

:::question type="NAT" question="Given the polynomial $P(x)=x^3-6x^2+11x-6$ with roots $r_1<r_2<r_3$, find the value of $\log_{r_1+r_2}(r_3^2)$." answer="2" hint="Find the roots first." solution="Test integer roots:

So

Hence

Now

Therefore
"
:::

:::question type="NAT" question="Solve for $x$: $x^{\log_x(x^2-3x+3)}=\log_2 8$. Provide the valid integer solution." answer="3" hint="Simplify both sides and check the log base conditions." solution="Since

whenever $x>0,\ x\ne 1,\ A>0$, the left side becomes

Also

So



This gives $x=0$ or $x=3$.

But $x=0$ is invalid because the base of the logarithm must be positive and not equal to $1$.

Hence the valid integer solution is

"
:::

---

Final Chapter Takeaways

Polynomials and Logarithms

Overview

This chapter covers two high-yield topics for CMI: polynomials and logarithms. Focus on definitions, core theorems, standard transformations, and structural properties. ---

Part 1: Introduction to Polynomials

---

Key Theorems

1. Remainder and Factor Theorems

When $P(x)$ is divided by $(x-c)$:

• Remainder Theorem: The remainder is exactly $P(c)$.

• Factor Theorem: $(x-c)$ is a factor if and only if $P(c) = 0$.

2. Fundamental Theorem of Algebra

A polynomial of degree $n \ge 1$ has exactly $n$ complex roots (counting multiplicity).

• Conjugate Root Theorem: If $P(x)$ has real coefficients and $a+bi$ is a root, then $a-bi$ is also a root.

3. Vieta's Formulas

For $P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_0$ with roots $r_1, r_2, \dots, r_n$:

• $\sum r_i = -\dfrac{a_{n-1}}{a_n}$

• $\sum_{i<j} r_i r_j = \dfrac{a_{n-2}}{a_n}$

• $r_1 r_2 \dots r_n = (-1)^n \dfrac{a_0}{a_n}$

4. Integer Coefficient Property

If $P(x)$ has integer coefficients, then for any distinct integers $a$ and $b$: $(a-b) \mid (P(a) - P(b))$ ---

Part 2: Logarithms

Core Properties

• Product: $\log_b(MN) = \log_b M + \log_b N$

• Quotient: $\log_b \left(\dfrac{M}{N}\right) = \log_b M - \log_b N$

• Power: $\log_b(M^k) = k \log_b M$

• Change of Base: $\log_b a = \dfrac{\log_c a}{\log_c b}$

---

Practice Questions (Conceptual)

:::question type="MCQ" question="Let $P(x)$ be a polynomial with integer coefficients such that $P(1) = 3$ and $P(3) = 10$. Which of the following is true?" options=["A) Such a polynomial exists.","B) Such a polynomial cannot exist.","C) Such a polynomial must be quadratic.","D) $P(x)$ must have a root in $(1, 3)$."] answer="B" hint="Use the integer divisibility property." solution="By the integer coefficient property, $(3-1)$ must divide $P(3) - P(1)$. $3 - 1 = 2$ and $P(3) - P(1) = 10 - 3 = 7$. Since $2$ does not divide $7$, no such polynomial exists. Correct option is $\boxed{B}$." ::: :::question type="MSQ" question="Let $P(x)$ be a polynomial of degree 4 with real coefficients. If $i$ and $1+i$ are roots, which of the following are also roots?" options=["A) $-i$","B) $1-i$","C) $0$","D) $2$"] answer="A,B" hint="Use the Conjugate Root Theorem." solution="Since the coefficients are real, complex roots must occur in conjugate pairs.

The conjugate of $i$ is $-i$.

The conjugate of $1+i$ is $1-i$.

Since the degree is 4, these four roots ($i, -i, 1+i, 1-i$) are the only roots. Correct options are $\boxed{A, B}$." :::

Practice Exercises: Polynomials & Logarithms

Conceptual & Theoretical Challenges

:::question type="MSQ" question="Let $P(x)$ be a polynomial of degree $n \ge 1$ with real coefficients. Which of the following statements must be true?" options=["A) If $n$ is odd, $P(x)$ must have at least one real root.","B) If $n$ is even, $P(x)$ must have at least one real root.","C) If $a+bi$ ($b \ne 0$) is a root, then $a-bi$ is also a root.","D) $P(x)$ can have at most $n$ distinct complex roots."] answer="A,C,D" hint="Consider the Intermediate Value Theorem for odd degrees and the Conjugate Root Theorem." solution="

• A is True: For odd degree polynomials, as $x \to \infty$, $P(x) \to \infty$ and as $x \to -\infty$, $P(x) \to -\infty$ (or vice versa). By IVT, it must cross the x-axis.

• B is False: Example: $P(x) = x^2 + 1$ has no real roots.

• C is True: Non-real roots of polynomials with real coefficients always occur in conjugate pairs.

• D is True: By the Fundamental Theorem of Algebra, a degree $n$ polynomial has exactly $n$ roots counting multiplicity, thus at most $n$ distinct roots.

Correct options are $\boxed{A, C, D}$." ::: :::question type="MSQ" question="Consider the equation $\log_x(y) + \log_y(x) = 2$ for real numbers $x, y$. Which of the following conditions must be satisfied?" options=["A) $x > 0$ and $y > 0$","B) $x \ne 1$ and $y \ne 1$","C) $x = y$","D) $x = \dfrac{1}{y}$"] answer="A,B,C" hint="Use the property $\log_y x = \dfrac{1}{\log_x y}$ and the AM-GM inequality." solution=" For the logarithms to be defined, we need $x, y > 0$ and $x, y \ne 1$ (Options A and B). Let $t = \log_x y$. Then the equation is $t + \dfrac{1}{t} = 2$. This implies $t^2 - 2t + 1 = 0$, so $(t-1)^2 = 0 \implies t = 1$. $\log_x y = 1 \implies x^1 = y$, so $x = y$. Note: While $x = 1/y$ satisfies $t + 1/t = -2$ if $t = -1$, it does not satisfy this specific equation. Correct options are $\boxed{A, B, C}$." ::: :::question type="SUB" question="Let $P(x)$ be a polynomial with integer coefficients such that $P(0) = n$ and $P(1) = m$, where $n$ and $m$ are integers. If $P(k) = 0$ for some integer $k$, prove that $k$ must divide $n$ and $(k-1)$ must divide $m$." answer="Proof based on divisibility property $(a-b) \mid (P(a)-P(b))$." hint="Substitute $a=k$ and $b=0$, then $a=k$ and $b=1$ into the integer coefficient property." solution=" By the integer coefficient property, for any integers $a, b$, $(a-b)$ divides $P(a) - P(b)$.

Let $a = k$ and $b = 0$. Then $(k-0)$ divides $P(k) - P(0)$.

Since $P(k) = 0$ and $P(0) = n$, we have $k \mid (0 - n)$, which means $k \mid n$.

Let $a = k$ and $b = 1$. Then $(k-1)$ divides $P(k) - P(1)$.

Since $P(k) = 0$ and $P(1) = m$, we have $(k-1) \mid (0 - m)$, which means $(k-1) \mid m$. This concludes the proof." ::: :::question type="SUB" question="Find all real values of $x$ that satisfy the equation: $\log_2(x^2 - 3x + 2) = \log_2(x-1) + 1$." answer="x = 3" hint="Check the domain of the logarithms first." solution=" Step 1: Domain Restrictions.

• $x^2 - 3x + 2 > 0 \implies (x-1)(x-2) > 0 \implies x < 1$ or $x > 2$.

• $x-1 > 0 \implies x > 1$.

Combining these, the valid domain is $x > 2$. Step 2: Solve the equation. Using $\log_b A + \log_b B = \log_b(AB)$: $\log_2(x^2 - 3x + 2) = \log_2(x-1) + \log_2(2)$ $\log_2(x^2 - 3x + 2) = \log_2(2(x-1))$ $x^2 - 3x + 2 = 2x - 2$ $x^2 - 5x + 4 = 0$ $(x-1)(x-4) = 0 \implies x = 1$ or $x = 4$. Step 3: Verify against domain. $x = 1$ is not in the domain ($x > 2$). $x = 4$ is in the domain. Final solution: $\boxed{x = 4}$." :::

Polynomials and Logarithms: Unified Theory

1. Polynomial Theory

A polynomial $P(x)$ of degree $n$ is defined by the coefficients $a_0, \dots, a_n$. The structure of its roots is governed by several core principles:

Fundamental Operations & Theorems

• Division Algorithm: $P(x) = D(x)Q(x) + R(x)$ where $\deg(R) < \deg(D)$.

• Remainder Theorem: Dividing $P(x)$ by $(x-c)$ yields a remainder of $P(c)$.

• Factor Theorem: $(x-c)$ is a factor $\iff P(c) = 0$.

• Fundamental Theorem of Algebra: A degree $n$ polynomial has exactly $n$ complex roots (counting multiplicities).

• Conjugate Root Theorem: For real coefficients, non-real roots must appear as $a \pm bi$.

Root-Coefficient Relationships (Vieta's Formulas)

For $P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_0$:

Sum of Roots: $e_1 = -\dfrac{a_{n-1}}{a_n}$

Sum of Roots taken two at a time: $e_2 = \dfrac{a_{n-2}}{a_n}$

Product of Roots: $e_n = (-1)^n \dfrac{a_0}{a_n}$

Advanced Structural Properties

• Rational Root Theorem: For integer coefficients, a rational root $\dfrac{p}{q}$ requires $p \mid a_0$ and $q \mid a_n$.

• Integer Divisibility Property: For integer $a, b$, $(a-b) \mid (P(a) - P(b))$.

• Intermediate Value Theorem (IVT): If $P(a)P(b) < 0$, at least one real root exists in $(a, b)$.

---

2. Logarithmic Theory

Logarithms serve as the inverse of exponentiation, critical for scaling data and solving growth equations.

Definitions & Domain

• Identity: $\log_b N = x \iff b^x = N$.

• Constraints: $N > 0$ (Argument), $b > 0$ and $b \ne 1$ (Base).

Operational Laws

• Product: $\log_b(MN) = \log_b M + \log_b N$

• Quotient: $\log_b\left(\dfrac{M}{N}\right) = \log_b M - \log_b N$

• Change of Base: $\log_b a = \dfrac{\log_c a}{\log_c b}$

• Base Power: $\log_{b^k} a = \dfrac{1}{k} \log_b a$

---

3. Graded Conceptual Challenges (Simple to Hard)

:::question type="MSQ" question="Let $P(x)$ be a polynomial of degree 4 with real coefficients. If $i$ and $1+i$ are roots, which statements are true?" options=["A) $-i$ is a root.","B) $1-i$ is a root.","C) $P(x)$ has no real roots.","D) $P(x)$ must have a constant term of 2 if it is monic."] answer="A,B,C,D" hint="Use the Conjugate Root Theorem and Vieta's for the constant term." solution="

By Conjugate Root Theorem: $-i$ and $1-i$ are roots. (A, B true).

Total roots = 4. All four are complex, so no real roots exist. (C true).

Product of roots $r_1r_2r_3r_4 = (i)(-i)(1+i)(1-i) = (1)(2) = 2$.

For a monic degree 4 polynomial, $a_0 = (-1)^4 (\text{Product}) = 2$. (D true).

Correct options: $\boxed{A, B, C, D}$." ::: :::question type="SUB" question="Prove that there is no polynomial $P(x)$ with integer coefficients such that $P(7) = 5$ and $P(15) = 9$." answer="No such polynomial exists." hint="Apply the property $(a-b) \mid (P(a) - P(b))$." solution=" If $P(x)$ has integer coefficients, then for integers $a=15$ and $b=7$: $(15-7)$ must divide $P(15) - P(7)$. $15 - 7 = 8$ and $P(15) - P(7) = 9 - 5 = 4$. Since $8$ does not divide $4$, no such polynomial exists. " ::: :::question type="SUB" question="Solve for real $x$: $\log_x(2x^2 - 5x + 3) = 2$." answer="x = 3" hint="Check base and argument constraints first." solution=" Step 1: Domain.

• Base: $x > 0, x \ne 1$.

• Argument: $2x^2 - 5x + 3 > 0 \implies (2x-3)(x-1) > 0 \implies x < 1$ or $x > 1.5$.

Combined: $x > 1.5$. Step 2: Solve. $x^2 = 2x^2 - 5x + 3 \implies x^2 - 5x + 3 = 0$. Using quadratic formula: $x = \dfrac{5 \pm \sqrt{13}}{2}$.

• $x_1 = \dfrac{5 + \sqrt{13}}{2} \approx 4.3$ (Valid: $> 1.5$)
  • $x_2 = \dfrac{5 - \sqrt{13}}{2} \approx 0.7$ (Invalid: $< 1.5$) $\boxed{x = \dfrac{5 + \sqrt{13}}{2}}$ " :::

    Polynomials and Logarithms: Master Theory

    1. Polynomial Foundations

    A polynomial $P(x)$ of degree $n$ is defined by the sum $\sum_{i=0}^{n} a_i x^i$. Its behavior is dictated by the coefficients $a_i$.

    Core Theorems for Problem Solving

    • Remainder Theorem: Dividing $P(x)$ by $(x-c)$ yields a remainder of $P(c)$.
    • Factor Theorem: $(x-c)$ is a factor of $P(x)$ if and only if $P(c) = 0$.
    • Conjugate Root Theorem: If $P(x)$ has real coefficients and $a+bi$ is a root, then $a-bi$ is also a root.
    • Fundamental Theorem of Algebra: Every degree $n$ polynomial has exactly $n$ complex roots, counting multiplicities.

    Vieta's Formulas (Symmetric Sums)

    For $a_n x^n + a_{n-1} x^{n-1} + \dots + a_0 = 0$ with roots $r_1, r_2, \dots, r_n$:
    • Sum of roots: $\sum r_i = -\dfrac{a_{n-1}}{a_n}$
    • Pairwise sum: $\sum_{i < j} r_i r_j = \dfrac{a_{n-2}}{a_n}$
    • Product of roots: $r_1 r_2 \dots r_n = (-1)^n \dfrac{a_0}{a_n}$

    High-Level Analysis

    • Intermediate Value Theorem (IVT): If $P(x)$ is real-valued and $P(a)P(b) < 0$, then $P(x)$ has at least one real root in $(a, b)$.
    • Integer Property: For $P(x) \in \mathbb{Z}[x]$, then for distinct integers $a, b$, the difference $(a-b)$ must divide $P(a) - P(b)$.
    ---

    2. Logarithmic Theory

    Logarithms transform multiplicative relationships into additive ones, defined as $\log_b N = x \iff b^x = N$.

    Domain & Identity

    • Constraints: Argument $N > 0$; Base $b > 0, b \ne 1$.
    • Basic Laws:
    - $\log_b(MN) = \log_b M + \log_b N$ - $\log_b(M/N) = \log_b M - \log_b N$ - $\log_b(M^k) = k \log_b M$
    • Base Change: $\log_b a = \dfrac{\log_c a}{\log_c b}$ and $\log_{b^k} a = \dfrac{1}{k} \log_b a$.
    ---

    3. Conceptual Challenges (MSQ & SUB)

    :::question type="MSQ" question="Let $P(x)$ be a monic polynomial with integer coefficients. Suppose $P(0) = 2025$. Which of the following statements must be true?" options=["A) Any rational root of $P(x)$ must be an integer.","B) If $r$ is an integer root, then $r$ must divide 2025.","C) $P(x)$ cannot have more than $\deg(P)$ real roots.","D) $P(x)$ must have at least one real root if its degree is odd."] answer="A,B,C,D" hint="Apply the Rational Root Theorem and the Fundamental Theorem of Algebra." solution="
    • A is True: For a monic polynomial (leading coefficient 1), the Rational Root Theorem states that any rational root $p/q$ must have $q \mid 1$, making it an integer.
    • B is True: The Rational Root Theorem also states $p \mid a_0$. Since $a_0 = P(0) = 2025$, any integer root must divide 2025.
    • C is True: A polynomial cannot have more real roots than its total degree.
    • D is True: By IVT, odd-degree polynomials with real coefficients must have at least one real root as the function ranges from $-\infty$ to $+\infty$.
    Correct options: $\boxed{A, B, C, D}$." ::: :::question type="MSQ" question="Consider the equation $\log_2(x) + \log_x(2) = \dfrac{5}{2}$. Which of the following are true?" options=["A) There are exactly two real solutions for $x$.","B) One of the solutions is $x = 4$.","C) One of the solutions is $x = \sqrt{2}$.","D) $x = 1$ is a valid solution."] answer="A,B,C" hint="Let $t = \log_2 x$ and solve the resulting quadratic." solution="
  • Constraint: $x > 0, x \ne 1$.
  • Let $t = \log_2 x$. Then $t + \dfrac{1}{t} = \dfrac{5}{2}$.
  • $2t^2 - 5t + 2 = 0 \implies (2t - 1)(t - 2) = 0$.
  • $t = 2 \implies \log_2 x = 2 \implies x = 4$.
  • $t = 1/2 \implies \log_2 x = 1/2 \implies x = \sqrt{2}$.
  • $x=1$ is invalid as it would make the base of $\log_x(2)$ undefined.
  Correct options: $\boxed{A, B, C}$." ::: :::question type="SUB" question="Let $P(x)$ be a polynomial with integer coefficients. If $P(1) = 1$ and $P(2) = 2$, can $P(5)$ be equal to 10? Justify your answer." answer="No" hint="Use the property $(a-b) \mid (P(a) - P(b))$." solution=" Using the integer coefficient property:
  • $(5-1) \mid (P(5) - P(1)) \implies 4 \mid (10 - 1) \implies 4 \mid 9$. This is False.
  • $(5-2) \mid (P(5) - P(2)) \implies 3 \mid (10 - 2) \implies 3 \mid 8$. This is False.
  Since these divisibility conditions are not met, no such polynomial exists. Answer: $\boxed{\text{No}}$." ::: :::question type="SUB" question="Solve the equation: $\log_3(x+1) + \log_3(x+3) = 1$ for all real $x$." answer="x = 0" hint="Combine logarithms and check the domain." solution=" Step 1: Domain. $x+1 > 0$ and $x+3 > 0 \implies x > -1$. Step 2: Solve. $\log_3((x+1)(x+3)) = 1$ $(x+1)(x+3) = 3^1$ $x^2 + 4x + 3 = 3$ $x^2 + 4x = 0 \implies x(x+4) = 0$. The solutions are $x = 0$ and $x = -4$. Step 3: Verify. $x = -4$ is outside the domain ($x > -1$). $x = 0$ is inside the domain. Final solution: $\boxed{x = 0}$." :::